Question

If $cos\alpha +cos\beta +cos\gamma =0=sin\alpha +sin\beta +sin\gamma$, then prove that $co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma =\frac{3}{2}=si{n}^2\alpha +si{n}^2\beta +si{n}^2\gamma$

Answer 1

$\cos \alpha +\cos \beta +\cos \gamma =0$

$\sin \alpha +\sin \beta +\sin \gamma =0$

$\cos \alpha +\cos \beta +\cos \gamma +\iota (\sin \alpha +\sin \beta +\sin \gamma )=0$

Let $\cos \alpha +\iota \left(\sin \alpha \right)=a$

$\cos \beta +\iota \left(\sin \beta \right)=b$

$\cos \gamma +\iota \left(\sin \gamma \right)=c$

So, $a+b+c=0-\left(1\right)$

${(a+b+c)}^2=0$

$\cos \alpha +\cos \beta +\cos \gamma -\iota (\sin \alpha +\sin \beta +\sin \gamma )=0-\iota \left(0\right)$

$=0$

So, $\cos \alpha -\iota \sin \alpha +\cos \beta -\iota \sin \beta +\cos \gamma -\iota \sin \gamma =0$

$\Rightarrow \frac{1}{\cos \alpha -\iota \sin \alpha }+\frac{1}{\cos \beta -\iota \sin \beta }+\frac{1}{\cos \gamma -\iota \sin \gamma }=0$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$

$\Rightarrow \frac{ab+bc+ca}{abc}=0$

$\Rightarrow ab+bc+ca=0$

As by $\left(1\right)$

${(a+b+c)}^2=a^2+b^2+c^2+2(ab+bc+ca)=0$

${(a+b+c)}^2=a^2+b^2+c^2+2\left(0\right)=0-\left(2\right)$

$\Rightarrow a^2+b^2+c^2=0$

$\Rightarrow \cos 2\alpha +\cos 2\beta +\cos 2\gamma +\iota (\sin 2\alpha +\sin 2\beta +\sin 2\gamma )=0$

$\Rightarrow \cos 2\alpha +\cos 2\beta +\cos 2\gamma =0$

$\Rightarrow 2{\cos }^2\alpha -1+2{\cos }^2\beta -1+2{\cos }^2\gamma -1=0$

$\Rightarrow {\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =\frac{3}{2}$

${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =3-({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma )=3-\frac{3}{2}$

$=\frac{3}{2}$